Perfect Square

Date: 10/26/2001 at 04:34:19
From: robin
Subject: perfect square
Hello Dr. Maths
Please help me in solving the following question:
If a and b are positive integers such that (1+ab) divides (a^2+b^2),
show that the integer (a^2+b^2) must be a perfect square.
---------
(1+ab)
Thanks.

Date: 10/26/2001 at 08:17:55
From: Doctor Floor
Subject: Re: Perfect square
Hi, Robin,
Thanks for writing.
First we note that for a = b = 1 indeed (a^2+b^2)/(1+ab) = 1 is a
perfect square. In other cases a and b can't be equal, so that we can
assume that a > b. Second we note that (a^2+b^2)/(1+ab) must be
positive.
Then, when 1+ab is a divisor of a^2+b^2, there must be a positive
integer N satisfying
a^2 + b^2
--------- = N
1 + ab
with a > b - except when a = b = 1, a case we have seen above (a can't
be equal to b except when a = b = 1). So we have
a^2 - (Nb)a + (b^2 - N) = 0.
This means that the quadratic equation
x^2 - (Nb)x + (b^2 - N) = 0
has solution x = a. The sum of the two solutions is Nb, so that the
second solution is x = Nb-a.
This brings us a second integer pair a' = (Nb-a), b' = b that
satisfies
(a')^2 + (b')^2
--------------- = N.
1 + a'b'
We show that a' < b' by writing the original equation in the form
Nb - a = (b^2 - N)/a, so that we have a' = (b^2 - N)/a. Now we derive
b(b-a) < 0 < N
b^2 - ab < N
b^2 - N < ab
(b^2 - N)/a < b
a' < b = b'
Repeating this process, we have a strictly decreasing sequence of
integers given by
s(0) = a,
s(1) = b,
s(k) = Ns(k-1) - s(k-2) (this generalizes a' = Nb-a)
satisfying
s(k)^2 + s(k-1)^2
----------------- = N.
1 + s(k)s(k-1)
The key is to show that this sequence must pass through 0, because if
s(j) = 0 for some integer j, then
s(j-1)^2 + s(j)^2
----------------- = s(j-1)^2 = N
1 + s(j)s(j-1)
and thus indeed N is a perfect square.
To prove the sequence passes through zero, suppose the sequence
doesn't. It follows that, since the sequence is strictly decreasing,
it must contain two x = s(n) and y = s(n+1) with opposite signs. Thus
(x^2 + y^2)/(1 + xy) must be either infinite (if xy = -1) or negative
(if xy < -1). But that contradicts N being a positive integer.
That completes the proof.
If you need more help, just write back.
Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/